Lời giải:
a) ĐKXĐ: \(x>0; x\neq 1\)
b)
\(A=\frac{\sqrt{x}(\sqrt{x}^3-1)}{x+\sqrt{x}+1}-\frac{2\sqrt{x}.\sqrt{x}+\sqrt{x}}{\sqrt{x}}+\frac{2(\sqrt{x}^2-1^2)}{\sqrt{x}-1}\)
\(=\frac{\sqrt{x}(\sqrt{x}-1)(x+\sqrt{x}+1)}{x+\sqrt{x}+1}-\frac{\sqrt{x}(2\sqrt{x}+1)}{\sqrt{x}}+\frac{2(\sqrt{x}-1)(\sqrt{x}+1)}{\sqrt{x}-1}\)
\(=\sqrt{x}(\sqrt{x}-1)-(2\sqrt{x}+1)+2(\sqrt{x}+1)\)
\(=x-\sqrt{x}-2\sqrt{x}-1+2\sqrt{x}+2=x-\sqrt{x}+1\)
c)
Ta có:
\(A=x-\sqrt{x}+1=(\sqrt{x})^2-2\sqrt{x}.\frac{1}{2}+(\frac{1}{2})^2+\frac{3}{4}\)
\(=(\sqrt{x}-\frac{1}{2})^2+\frac{3}{4}\geq \frac{3}{4}\) với mọi $x>0; x\neq 1$
Vậy GTNN của $A$ là $\frac{3}{4}$. Dấu "=" xảy ra khi \(\sqrt{x}-\frac{1}{2}=0\Leftrightarrow x=\frac{1}{4}\)